1201. Ugly Number III

问题：

给定a，b，c

求出第n个能被a或b或c整除的正整数。（这种数也被称为：Ugly Number）

Example 1:
Input: n = 3, a = 2, b = 3, c = 5
Output: 4
Explanation: The ugly numbers are 2, 3, 4, 5, 6, 8, 9, 10... The 3rd is 4.

Example 2:
Input: n = 4, a = 2, b = 3, c = 4
Output: 6
Explanation: The ugly numbers are 2, 3, 4, 6, 8, 9, 10, 12... The 4th is 6.

Example 3:
Input: n = 5, a = 2, b = 11, c = 13
Output: 10
Explanation: The ugly numbers are 2, 4, 6, 8, 10, 11, 12, 13... The 5th is 10.

Example 4:
Input: n = 1000000000, a = 2, b = 217983653, c = 336916467
Output: 1999999984
 
Constraints:
1 <= n, a, b, c <= 10^9
1 <= a * b * c <= 10^18
It's guaranteed that the result will be in range [1, 2 * 10^9]

解法：二分查找(Binary Search)

找到第一个正整数，使得其之前有n个，能被a或b或c整除的正整数。

求解对象范围：正整数：

• 最小：l：1
• 最大：r：INT_MAX

求解，m以内有多少个UglyNumber。

long getCount(long a, long b, long c, long m)

⚠️ 注意：

m以内有m/a个数能被a整除。

m以内有m/lcm(a,b)个数能被a且b整除。（lcm(a,b)为a和b的最小公倍数）

那么有，m以内有 m/a+m/b-m/lcm(a,b) 个数能被a或b整除。

能被a整除的 + 能被b整除的 - 能被a且b整除的（前面+算了两遍，这里需要减去一遍）

因此，对本问题则有：

 要求的 能被a或b或c整除的 个数：

= 分别能被a,b,c整除的个数之和 - 分别能被a&&b,a&&c,b&&c整除的个数之和 + 能被a&&b&&c整除的个数之和

代码参考：

class Solution {
public:
    long getCount(long a, long b, long c, long m) {
        // 获取m以内，有多少个数，能被a或b或c整除（Ugly number）
        long ab_count = m/lcm(a,b);
        long ac_count = m/lcm(a,c);
        long bc_count = m/lcm(b,c);
        long abc_count = m/lcm(lcm(a,b),c);
        return m/a + m/b + m/c - ab_count - ac_count - bc_count + abc_count;
    }
    int nthUglyNumber(int n, long a, long b, long c) {
        long l = 1, r = INT_MAX;
        while(l<r) {
            long m = l+(r-l)/2;
            if(getCount(a,b,c,m)>=n) {
                r = m;
            }else{
                l = m+1;
            }
        }
        return l;
    }
};